Đáp án:

\(\int_{0}^{1} [f'(x) \cdot (f(x))^2 + 1] dx = 2 \int_{0}^{1} f(x) \cdot \sqrt{f'(x)} dx\)

\(\Leftrightarrow \int_{0}^{1} (\sqrt{f'(x)} \cdot f(x) - 1)^2 dx = 0 \Leftrightarrow \sqrt{f'(x)} \cdot f(x) = 1\)

\(\Leftrightarrow f'(x) \cdot (f(x))^2 = 1\)

\(\Rightarrow \int f'(x) \cdot (f(x))^2 dx = \int dx \Rightarrow \frac{(f(x))^3}{3} = x + C\)

\(\Rightarrow(f(x))^3 = 3x + 3C\)

\(f(0) = 2 \Rightarrow 3C = 8\)

\((f(x))^3 = 3x + 8 \Rightarrow \int_{0}^{1} (f(x))^3 dx = \frac{3x^2}{2} + 8x \big|_{0}^{1} = \frac{19}{2} \Rightarrow \boxed{D}\)

page36

Đáp án:

Vì \(f'(x) \geq 0\), \(f(x) > 0\), \(\forall x \in (0, +\infty)\):  

\(\Rightarrow f'(x) = \sqrt{(x+1)f(x)}.\)

\(\Rightarrow\frac{f'(x)}{\sqrt{f(x)}} = \sqrt{x+1} \quad \Rightarrow \quad \int \frac{f'(x)}{\sqrt{f(x)}} dx = \int \sqrt{x+1} dx.\)

\(\Rightarrow2\sqrt{f(x)} = \frac{2}{3} \sqrt{(x+1)}^{3} + C.\)

\(\Rightarrow \sqrt{f(x)} = \frac{1}{3}\sqrt{(x+1)}^{3} + C.\)

Với \(f(2) = 3 \Rightarrow \quad C = 0.\)

\(\Rightarrow f(x) = \frac{1}{9}(x+1)^{3}.\)

 \( \Rightarrow f(8) = 81 \Rightarrow \boxed{81}.\)

page37

Đáp án:

\(x^2 (f(x))^2 + (2x - 1)f(x) = x f'(x) - 1\)

\(\Leftrightarrow x^2 (f(x))^2 + 2x f(x) + 1 = x f'(x) + f(x)\)

\(\Leftrightarrow (x f(x) + 1)^2 = (x f(x))'\)

\(\Leftrightarrow \frac{ (x f(x))'}{x f(x) + 1)^2} = 1 \Leftrightarrow \frac{(x f(x) + 1)'}{(x f(x) + 1)^2} = 1\)

\(\Leftrightarrow -\frac{1}{x f(x) + 1} = x + C\)

\(f(1) = -2 \Rightarrow C = 0\)

\(\Rightarrow x f(x) + 1 = -\frac{1}{x} \Rightarrow f(x) = \frac{1}{x} \left(-1 - \frac{1}{x}\right) = -\frac{1}{x} - \frac{1}{x^2}\)

\(\int_{1}^{2} f(x) dx = \int_{1}^{2} \left(-\frac{1}{x} - \frac{1}{x^2}\right) dx = -\ln{x} + \frac{1}{x} \Big|_{1}^{2}\)

\(= -\ln{2} + \frac{1}{2} - 1 = -\frac{1}{2} - \ln{2}\Rightarrow \boxed{B}\)

Gợi ý: \( (x f(x))' = f(x) + x f'(x)\)

page38

Đáp án:

\(2x(1 + f(x)) = [f'(x)]^3 \Leftrightarrow f'(x) = \sqrt[3]{2x(1 + f(x))}\)

\(\Rightarrow \frac{f'(x)}{\sqrt[3]{1 + f(x)}} = \sqrt[3]{2x} \)

\(\Rightarrow \quad \int f'(x)(1 + f(x))^{-\frac{1}{3}} dx = \sqrt[3]{2} \int x^{\frac{1}{3}} dx\)

\(\Rightarrow \frac{3}{2}(1 + f(x))^{\frac{2}{3}} = \sqrt[3]{2}\frac{3}{4} x^{\frac{4}{3}} + C\)

\(\Rightarrow (1 + f(x))^{\frac{2}{3}} = \frac{\sqrt[3]{2}}{2} x^{\frac{4}{3}} + C\)

\(f(0) = -1 \Rightarrow C = 0\)

\(\Rightarrow (1 + f(x))^2 = \frac{1}{4} x^4 \quad \Rightarrow \quad 1+ f(x) = \frac{1}{2}x^2 \)

\(\Rightarrow f(x) = \frac{1}{2}x^2 -1 \)

\(\int_{0}^{1} f(x) dx = \int_{0}^{1} \left(\frac{1}{2}x^2 - 1\right) dx = -\frac{5}{6}\)

page39

Đáp án:

\(x(1 + 2f(x)) = (f'(x))^2\)

\(\Rightarrow \frac{(f'(x))^2}{1 + 2f(x)} = x \quad \Rightarrow \quad \frac{f'(x)}{\sqrt{1 + 2f(x)}} = \sqrt{x}, \, \forall x \in [1, 4]\)

\(\Rightarrow \int \frac{f'(x)}{\sqrt{1 + 2f(x)}} dx = \int \sqrt{x} dx\)

\(\Rightarrow \sqrt{1 + 2f(x)} = \frac{2}{3} x \sqrt{x} + C\)

\(f(1) = \frac{3}{2} \Rightarrow 2 = \frac{2}{3} \ + C \Rightarrow C = 2 - \frac{2}{3} = \frac{4}{3}\)

\(\Rightarrow 1 + 2f(x) = \left(\frac{2}{3} x \sqrt{x} + \frac{4}{3} \right)^2 = \frac{4}{9}x^3 + \frac{16}{9}x  \sqrt{x} + \frac{16}{9}\)

\(\Rightarrow f(x) = \frac{1}{2}\left[\frac{4}{9}x^3 + \frac{16}{9}x  \sqrt{x}+ \frac{7}{9}\right]\)

\(\Rightarrow \int_{1}^{4} f(x) dx  = \frac{1186}{45} \Rightarrow \boxed{A}\)

page40